5(x+y)-3(y/x) plz help me

Mathematics

1 answer:

Contact [7]3 years ago

4 0

Answer:

$\large\boxed{5(x+y)-3\left(\dfrac{y}{x}\right)=5x+5y-\dfrac{3y}{x}}$

Step-by-step explanation:

$5(x+y)\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\\\=5x+5x\\\\3\left(\dfrac{y}{x}\right)=\dfrac{3y}{x}\\\\5(x+y)-3\left(\dfrac{y}{x}\right)=5x+5y-\dfrac{3y}{x}$

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If 1 is added to a number and the sum is tripled, the result is 17 more than the number. Find the number

bearhunter [10]

Form an equation. Start by representing the number you want to find as X. Now add 1.
X + 1
Now we need to triple it. That basically means multiply X+1 by 3. You can write that as:
3 (X + 1)
Now we write out that the result is 17 more than the number we want to find. So set an equals sign up
3 (X + 1) =
And now write X + 17
3 (X + 1) = X + 17

Now we solve for X.
Firstly, let's get rid of the X on the right.
Remember, if we want to get rid of a number of variable on one side, we need to use opposites, and do the same to the other side of the equation.
In this case, we have a positive X on the right. To get rid of that, we subtract X from both sides. This will leave us with:
2x + 3 = 17
Now let's get rid of the +3 on the left. To do this, we subtract 3 from both sides of the equation.
2x = 17 - 3
2x = 14
Now we just need to get X by itself. The 2 in front of the X multiplies X by 2. Therefore, to get X on its own, we need to divide 2x by 2, which just gives us 2.
We also need to divide the other side by 2, because if we do something on one side, we need to do something on the other too. So divide 14 by 2, and we get 7.
X = 7

But before we celebrate, let's double check the answer. Add 1 to 7. This gives us 8. Triple 8. This is 24. This should be 17 more than 7. So go, 24 - 17. And whaddya know - it's 7!

4 0

3 years ago

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$